Title says it. Combine 8 decks, shuffle, deal three cards. What are the odds that these three cards make a straight. Not just looking for the answer as I believe I found that online. Really I'd like to understand what I am doing wrong. My logic:
$\frac{11}{13}$ of the time I'll get Ace or 3-Q for card 1. Card 2 can therefore go any direction and be within 2 values to keep the straight going. This is $\frac{128}{415}$. Card 3 only has 1 possible value which has odds $\frac{32}{414}$.
The other $\frac{2}{13}$ of the time I'll get a 2 or King for card 1. There are now two paths to take. Path 1: $\frac{128}{415}$ I'll get a second card within 2 values towards the center, and $\frac{32}{414}$ of the time card 3 will complete. Path 2: $\frac{32}{415}$ of the time my second card is an Ace and it's still $\frac{32}{414}$ to complete the straight with card 3.
This makes:
$(\frac{11}{13} \cdot \frac{128}{415} \cdot \frac{32}{414}) + (\frac{2}{13} \cdot \frac{128}{415} \cdot \frac{32}{414}) + (\frac{2}{13} \cdot \frac{32}{415} \cdot \frac{32}{414}) = \frac{3072}{12085} = 0.024757… $
However an online poker site I found gives the odds at $0.030947$ and a simulator I wrote to play 10,000,000 full shoes of 8 decks of cards gives $0.030549$ which is closer to the online site but far enough off that I must have a coding bug along with my actual hand math bug.
Best Answer
"Card 3 only has 1 possible value" is not true if, for example, you draw Q-K, so you are missing some possibilities there.
It is easier to count the number of unordered three-card hands that qualify, and divide that by the total number of unordered three-card hands.
There are 12 possible ranks for the straight: A-2-3 through Q-K-A, and there are 32 cards available for each rank within the straight. This gives a total of $12\times32^3=393216$ straights, but we need to exclude straight flushes. There are $12\times4\times8^3=24576$ of those, so the number of straights is $393216-24576=368640$
The total number of unordered three-card hands is $\binom{416}{3}=11912160$ and the probability of drawing a straight is therefore:$$\frac{368640}{11912160}\approx0.03094653$$